ATAR Notes: Forum

VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: /0 on April 23, 2009, 01:32:31 pm

Title: Inequality
Post by: /0 on April 23, 2009, 01:32:31 pm: Prove $x+\frac{1}{x} \geq 2$ for $x > 0$ WITHOUT CALCULUS
Title: Re: Inequality
Post by: kamil9876 on April 23, 2009, 03:16:30 pm: You can use the method i used for Ahmad's problem:
http://vcenotes.com/forum/index.php/topic,2398.msg140527.html#msg140527

checkout the problem in my second post of that thread, it has a similair strategy too :)

Your problem:

It's true for x=1. if increase x from 1 to 1+a:

then the expression: $x+\frac{1}{x}$ becomes:

$1+a + \frac{1}{1+a}$
$=2+a-\frac{a}{a+1}$ (1)

obviously -1<a since x is positive. If a was positive then $a>\frac{a}{a+1}$ [/tex] and so $a-\frac{a}{a+1}$
is positive which means that expression 1 is greater than two as required. However if -1<a<0 then let a=-b and so 0<b<1.

$a-\frac{a}{a+1}=\frac{b}{1-b}-b$

But obviously $\frac{b}{1-b}$ has a larger modulus than b. and so $a-\frac{a}{a+1}=\frac{b}{1-b}-b$ is positive in this case too.
Title: Re: Inequality
Post by: /0 on April 23, 2009, 07:31:50 pm: nice! thanks kamil
Title: Re: Inequality
Post by: Mao on April 23, 2009, 09:25:09 pm: $x + \frac{1}{x} \ge 2$

$\implies x^2 + 1 \ge 2x, \; x>0$

$\implies x^2 - 2x + 1 \ge 0,\; x>0$

$\implies (x-1)^2 \ge 0$ , which is true

$\therefore x + \frac{1}{x} \ge 2, \; x>0$
Title: Re: Inequality
Post by: TrueTears on April 23, 2009, 09:29:14 pm: Interesting methods lol
Title: Re: Inequality
Post by: TonyHem on April 23, 2009, 09:31:27 pm: Quote from: Mao on April 23, 2009, 09:25:09 pm
$x + \frac{1}{x} \ge 2$

$\implies x^2 + 1 \ge 2x, \; x>0$

$\implies x^2 - 2x + 1 \ge 0,\; x>0$

$\implies (x-1)^2 \ge 0$ , which is true

$\therefore x + \frac{1}{x} \ge 2, \; x>0$

omg I had that too!
haha